How should I expand 1/(1+x)^n around x=0?

  • Thread starter liyz06
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In summary, to expand 1/(1+x)^n around x=0 using Taylor series, you can use the general formula ## \displaystyle f(x)=f(0)+\frac {f'(0)}{1!} (x)+ \frac{f''(0)}{2!} (x)^2+\frac{f^{(3)}(0)}{3!}(x)^3+ \cdots \ . ##
  • #1
liyz06
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Homework Statement



How should I expand 1/(1+x)^n around x=0?

Homework Equations


The Attempt at a Solution

 
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  • #2
Did you try a Taylor expansion?
What did you get? Where did you run into problems?
 
  • #3
mfb said:
Did you try a Taylor expansion?
What did you get? Where did you run into problems?

I know that (1+x)^n could be expanded easily by binomial theorem, but what I need here is to expand (1+x)^-n into polynomial form, not the reciprocal of a polynomial
 
  • #4
I don't see how your answer is related to my post.
You can just calculate the Taylor expansion.
 
  • #5
liyz06 said:

Homework Statement



How should I expand 1/(1+x)^n around x=0?

Homework Equations



The Attempt at a Solution

Are you familiar with Taylor Series ?

The Taylor expansion for a function, f(x), expanded about x = a is:

## \displaystyle f(x)=f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots \ . ##

So, expanding about x = 0 gives:

## \displaystyle f(x)=f(0)+\frac {f'(0)}{1!} (x)+ \frac{f''(0)}{2!} (x)^2+\frac{f^{(3)}(0)}{3!}(x)^3+ \cdots \ . ##
 

FAQ: How should I expand 1/(1+x)^n around x=0?

What is the formula for expanding 1/(1+x)^n around x=0?

The general formula for expanding 1/(1+x)^n around x=0 is (1+x)^(-n) = 1 - nx + (n(n+1)/2!) x^2 - (n(n+1)(n+2)/3!) x^3 + ... This is known as the Binomial Series expansion.

How do I know when to stop expanding 1/(1+x)^n around x=0?

It is recommended to stop expanding when the term involving x^k is smaller than the desired level of accuracy. This means that we stop expanding when |(n(n+1)...(n+k-1)/k!) x^k| < ε, where ε is the desired level of accuracy.

Can I use this expansion for any value of n?

The expansion for 1/(1+x)^n around x=0 is valid for any real value of n. However, it may not converge for values of n that are negative or non-integer.

What is the significance of expanding 1/(1+x)^n around x=0?

Expanding 1/(1+x)^n around x=0 allows us to approximate the value of the function for values of x near 0. This can be helpful in simplifying complex expressions and solving differential equations.

Are there any other methods for expanding 1/(1+x)^n around x=0?

Yes, there are other methods such as using the Newton's Binomial Theorem or using the Taylor Series expansion for 1/(1+x). However, the Binomial Series expansion is the most commonly used method for expanding 1/(1+x)^n around x=0.

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